Cardano’s derivation of the cubic formula

To solve the cubic polynomial equation $x^3+a{x}^2+bx+c=0$ for $x$, the first step is to apply the Tchirnhaus transformation $x=y-\frac{a}{3}$. This reduces the equation to $y^3+py+q=0$, where

$p$ $=$ $b-{\displaystyle \frac{a^2}{3}}$

$q$ $=$ $c-{\displaystyle \frac{ab}{3}}+{\displaystyle \frac{2a^3}{27}}$

The next step is to substitute $y=u-v$, to obtain

$${(u-v)}^3+p(u-v)+q=0$$ (1)

or, with the terms collected,

$$(q-(v^3-u^3))+(u-v)(p-3uv)=0$$ (2)

From equation (2), we see that if $u$ and $v$ are chosen so that $q=v^3-u^3$ and $p=3uv$, then $y=u-v$ will satisfy equation (1), and the cubic equation will be solved!

There remains the matter of solving $q=v^3-u^3$ and $p=3uv$ for $u$ and $v$. From the second equation, we get $v=p/(3u)$, and substituting this $v$ into the first equation yields

$$q=\frac{p^3}{{(3u)}^3}-u^3$$

which is a quadratic equation in $u^3$. Solving for $u^3$ using the quadratic formula, we get

$u^3$ $=$ ${\displaystyle \frac{-27q+\sqrt{108p^3+729q^2}}{54}}={\displaystyle \frac{-9q+\sqrt{12p^3+81q^2}}{18}}$

$v^3$ $=$ ${\displaystyle \frac{27q+\sqrt{108p^3+729q^2}}{54}}={\displaystyle \frac{9q+\sqrt{12p^3+81q^2}}{18}}$

Using these values for $u$ and $v$, you can back–substitute $y=u-v$, $p=b-a^2/3$, $q=c-ab/3+2a^3/27$, and $x=y-a/3$ to get the expression for the first root $r_1$ in the cubic formula. The second and third roots $r_2$ and $r_3$ are obtained by performing synthetic division using $r_1$, and using the quadratic formula on the remaining quadratic factor.

Title	Cardano’s derivation of the cubic formula
Canonical name	CardanosDerivationOfTheCubicFormula
Date of creation	2013-03-22 12:10:28
Last modified on	2013-03-22 12:10:28
Owner	djao (24)
Last modified by	djao (24)
Numerical id	12
Author	djao (24)
Entry type	Proof
Classification	msc 12D10
Related topic	FerrariCardanoDerivationOfTheQuarticFormula